貪心-HDU3348 coins(錢幣問題)

題目

傳送門: HDU-3348

“Yakexi, this is the best age!” Dong MW works hard and get high pay, he has many 1 Jiao and 5 Jiao banknotes(紙幣), some day he went to a bank and changes part of his money into 1 Yuan, 5 Yuan, 10 Yuan.(1 Yuan = 10 Jiao)
“Thanks to the best age, I can buy many things!” Now Dong MW has a book to buy, it costs P Jiao. He wonders how many banknotes at least,and how many banknotes at most he can use to buy this nice book. Dong MW is a bit strange, he doesn’t like to get the change, that is, he will give the bookseller exactly P Jiao.

input:

T(T<=100) in the first line, indicating the case number.
T lines with 6 integers each:
P a1 a5 a10 a50 a100
ai means number of i-Jiao banknotes.
All integers are smaller than 1000000.

output:

Two integers A,B for each case, A is the fewest number of banknotes to buy the book exactly, and B is the largest number to buy exactly.If Dong MW can’t buy the book with no change, output “-1 -1”.

Sample Input:

3
33 6 6 6 6 6
10 10 10 10 10 10
11 0 1 20 20 20

Sample Output:

6 9
1 10
-1 -1

題意

分別給出1，5，10，50，100角面值的紙幣張數，求湊出p角的最小張數和最大張數。

分析

• 最小：每次貪心選擇面值最大的紙幣來湊成p角，湊不成則輸出-1。
• 最大：用對立反面來求，用最小張數湊成它的反面，那麼剩餘張數就是湊成p的最大張數。設所有紙幣組成總價值sum和紙幣總數cnt，用上一點求湊成(sum-p)最小張數x，cnt-x即可。

代碼

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 1000006;
int t, p, a[5], b[5] = { 1,5,10,50,100 };
int mi, ma, sum, cnt, fp, ti;
int main() {
	scanf("%d", &t);
	while (t--) {
		scanf("%d%d%d%d%d%d", &p, &a[0], &a[1], &a[2], &a[3], &a[4]);
		mi = ma = sum = cnt = 0;
		for (int i = 0; i <= 4; i++) {
			sum += b[i] * a[i];//紙幣總價值
			cnt += a[i];//紙幣總數
		}
		fp = sum - p;
		for (int i = 4; i >= 0; i--) {
			ti = p / b[i];//最多換b[i]面值幾張
			if (ti > a[i])ti = a[i];//超過題目輸入張數
			mi += ti;
			p -= ti * b[i];
		}
		if (p) {
			printf("-1 -1\n");
			continue;
		}
		for (int i = 4; i >= 0; i--) {
			ti = fp / b[i];
			if (ti > a[i])ti = a[i];
			ma += ti;
			fp -= ti * b[i];
		}
		printf("%d %d\n", mi, cnt - ma);
	}
	return 0;
}

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